From the docs:
Many operations have an “in-place” version. The following functions provide a more primitive access to in-place operators than the usual syntax does; for example, the statement x += y is equivalent to x = operator.iadd(x, y). Another way to put it is to say that z = operator.iadd(x, y) is equivalent to the compound statement z = x; z += y.
Questions:
Why isn't operator.iadd(x, y)
equivalent to z = x; z += y
?
How does operator.iadd(x, y)
differ from operator.add(x, y)
?
Related question, but I'm not interested in Python class methods; just regular operators on built-in Python types.
Difference between == and is operators in PythonThe == operator helps us compare the equality of objects. The is operator helps us check whether different variables point towards a similar object in the memory. We use the == operator in Python when the values of both the operands are very much equal.
Definition - In-place operation is an operation that changes directly the content of a given linear algebra, vector, matrices(Tensor) without making a copy. The operators which helps to do the operation is called in-place operator.
Standard operating procedure means a formal written procedure offi- cially adopted by the plant owner or operator and available on a routine basis to those persons responsible for carrying out the procedure.
First, you need to understand the difference between __add__
and __iadd__
.
An object's __add__
method is regular addition: it takes two parameters, returns their sum, and doesn't modify either parameter.
An object's __iadd__
method also takes two parameters, but makes the change in-place, modifying the contents of the first parameter. Because this requires object mutation, immutable types (like the standard number types) shouldn't have an __iadd__
method.
a + b
uses __add__
. a += b
uses __iadd__
if it exists; if it doesn't, it emulates it via __add__
, as in tmp = a + b; a = tmp
. operator.add
and operator.iadd
differ in the same way.
To the other question: operator.iadd(x, y)
isn't equivalent to z = x; z += y
, because if no __iadd__
exists __add__
will be used instead. You need to assign the value to ensure that the result is stored in both cases: x = operator.iadd(x, y)
.
You can see this yourself easily enough:
import operator
a = 1
operator.iadd(a, 2)
# a is still 1, because ints don't have __iadd__; iadd returned 3
b = ['a']
operator.iadd(b, ['b'])
# lists do have __iadd__, so b is now ['a', 'b']
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With